What is the solution for y^{''}+2y^'-3y=x^2-1 ?

The solution for y^{''}+2y^'-3y=x^2-1 is y=c_{1}e^x+c_{2}e^{-3x}-(x^2)/3-(4x)/9-5/27

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y^{''}+2y^'-3y=x^2-1

y^{^{''}} + 2 y^{^{'}} - 3 y = x^{2} - 1

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{x ^{2}}{3} - \frac{4 x}{9} - \frac{5}{27}

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = x^{2} - 1

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{x ^{2}}{3} - \frac{4 x}{9} - \frac{5}{27}

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{x ^{2}}{3} - \frac{4 x}{9} - \frac{5}{27}

\frac{d ^{2} x}{d t ^{2}} + 16 x = 12 cos (9 t)

y^{^{''}} - 3 y^{^{'}} - 18 y = x e^{3 x}

y^{^{''}} + y = e^{- t} sin (e^{- t}) + cos (e^{- t})

y^{^{''}} + y = \frac{x}{∣ x ∣}

(D^{2} + 5 D + 6) y = x^{2} + 2 x

What is the solution for y^{''}+2y^'-3y=x^2-1 ?
The solution for y^{''}+2y^'-3y=x^2-1 is y=c_{1}e^x+c_{2}e^{-3x}-(x^2)/3-(4x)/9-5/27